JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · Parabola

For a focal chord of parabola $y^2=4ax$, the tangents at its endpoints meet on the directrix $x=-a$.

Solving $y=2x+a$ with $x=-a$: $y=2(-a)+a=-a$.

So the tangents meet at $(-a,-a)$.

The equation of chord of contact from $(-a,-a)$ to the parabola is:

$(-a)y=2a(x+(-a))$

$-ay=2ax-2a^2$

$2ax+ay-2a^2=0$

$2x+y-2a=0$

Substituting $y^2=4ax$ into the line equation $y=2a-2x$:

$y^2=4a\cdot \frac{2a-y}{2}=2a(2a-y)$

$y^2+2ay-4a^2=0$

For the two values of $y$, if we write $y=mx$ where $m$ is the slope from origin:

$m^2-2m-4=0$ (after dividing by $x^2$ and using $y=2a-2x$)

The length of focal chord with slope $m$ is $\frac{4a}{m^2}\sqrt{a(a-mc)(1+m^2)}$ where $c=2a$.

Using the formula for focal chord: $PQ=\frac{4a}{{\sin }^2\theta }$ where $\theta$ is the angle the chord makes.

For $m=-2$: $PQ=\frac{4a}{4}\sqrt{a\cdot 5a\cdot 5}=5a$